Question Number 226471 by Rojarani last updated on 30/Nov/25
$$\:{If},\:{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} \infty{xy}\: \\ $$$$\:\:{then}\:{prove}\:{that},\:\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \infty{xy} \\ $$
Commented by fantastic2 last updated on 30/Nov/25
Commented by fantastic2 last updated on 30/Nov/25
$${press}\:{on}\:{star}\:{icon}. \\ $$$${then}\:{in}\:{the}\:{last}\:{row}\:{you}\: \\ $$$${will}\:{see} \\ $$$$\propto \\ $$
Answered by som(math1967) last updated on 30/Nov/25
$$\:{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} ={kxy} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{xy}}=\frac{{k}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{xy}}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{xy}}=\frac{{k}+\mathrm{2}\sqrt{\mathrm{2}}}{{k}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\left({x}+\sqrt{\mathrm{2}}{y}\right)^{\mathrm{2}} }{\left({x}−\sqrt{\mathrm{2}}{y}\right)^{\mathrm{2}} }=\frac{{k}+\mathrm{2}\sqrt{\mathrm{2}}}{{k}−\mathrm{2}\sqrt{\mathrm{2}}}\:\:\bigstar \\ $$$$\Rightarrow\frac{{x}+\sqrt{\mathrm{2}}{y}}{{x}−\sqrt{\mathrm{2}}{y}}=\sqrt{\frac{{k}+\mathrm{2}\sqrt{\mathrm{2}}}{{k}−\mathrm{2}\sqrt{\mathrm{2}}}}={k}_{\mathrm{1}} \left({say}\right) \\ $$$$\Rightarrow\frac{{x}}{\:\sqrt{\mathrm{2}}{y}}=\frac{{k}_{\mathrm{1}} +\mathrm{1}}{{k}_{\mathrm{1}} −\mathrm{1}}\:\bigstar \\ $$$$\Rightarrow\frac{{x}}{{y}}=\sqrt{\mathrm{2}}×\frac{{k}_{\mathrm{1}} +\mathrm{1}}{{k}_{\mathrm{1}} −\mathrm{1}}={k}_{\mathrm{2}} \:\left({say}\right) \\ $$$$\therefore{x}={k}_{\mathrm{2}} {y} \\ $$$$\:{Now}\:\frac{\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{xy}}=\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} {k}_{\mathrm{2}} ^{\mathrm{2}} }{{x}^{\mathrm{2}} {k}_{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{x}^{\mathrm{2}} \left(\mathrm{2}+{k}_{\mathrm{2}} ^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} {k}_{\mathrm{2}} }=\frac{\mathrm{2}+{k}_{\mathrm{2}} ^{\mathrm{2}} }{{k}_{\mathrm{2}} }={constant} \\ $$$$\therefore\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \propto{xy}\: \\ $$$$\bigstar\:\boldsymbol{{using}}\:\boldsymbol{{componendo}}\:\boldsymbol{{and}} \\ $$$$\:\boldsymbol{{dividendo}} \\ $$
Answered by AgniMath last updated on 30/Nov/25
$${x}^{\mathrm{2}} \:+\:\mathrm{2}{y}^{\mathrm{2}} \:\propto\:{xy} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{y}^{\mathrm{2}} \:=\:{kxy} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:−\:{kxy}\:+\:\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$$${Divide}\:{both}\:{sides}\:{by}\:{y}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} \:−\:{k}.\left(\frac{{x}}{{y}}\right)\:+\:\mathrm{2}\:=\:\mathrm{0}\: \\ $$$$\Rightarrow\:\frac{{x}}{{y}}\:=\:\frac{{k}\:\pm\:\sqrt{{k}^{\mathrm{2}} \:−\:\mathrm{8}}}{\mathrm{2}}\:=\:{Constant} \\ $$$$\Rightarrow\:{x}\:\propto\:{y} \\ $$$$\Rightarrow\:{x}\:=\:{my} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} }{{xy}}\:=\:\frac{\mathrm{2}{m}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} }{{my}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}{m}^{\mathrm{2}} \:+\:\mathrm{1}}{{m}}\:=\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Constant} \\ $$$$\Rightarrow\:\mathrm{2}{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}\:} \:\propto\:{xy}\:\left({Proved}\right) \\ $$
Commented by AgniMath last updated on 30/Nov/25
$${Try}\:{this}\:{for}\:{practice} \\ $$$${If}\:{ax}\:+\:{by}\:\propto\:\sqrt{{xy}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${ax}^{\mathrm{2}} \:+\:{by}^{\mathrm{2}} \:\propto\:{xy}. \\ $$